$\lim \limits_{h\to 0} \frac{e^h-1}{h}=1$ and the relation with $\sin'(x)=\cos(x)$

$$e^x=\sum _{n=0}^{\infty } \frac{x^n}{n!}$$

\begin{align*}\lim_{h\to 0} \frac{ e^{(x+h)}-e^x}{h}=e^x\color{blue}{\lim_{h\to 0} \frac{e^h-1}{h}}=e^x\tag{1}\end{align*}

How obvious about the following limit?

I think it is obvious when using L'Hospital Rule, $\frac{0}{0}$

However we haven't had the derivative of $e^x$, so can one explain me a bit about this limit...a little tiny, but I stuck...

\begin{align*}\lim_{h\to 0} \frac{e^h-1}{h}=1\tag{2}\end{align*}

And why (1) contain the fact $\sin'(x)=\cos(x)$ and $\cos'(x)=-\sin(x)$

• How do you define $e^x$? – Seirios Jul 21 '13 at 13:42
• @Seirios Power Series – HyperGroups Jul 21 '13 at 13:43

1 Answer

Assuming that you know the series expansion of $e^{x}$, we have

$$e^{h} = 1 + h + \frac{h^{2}}{2!} + \cdots = 1 + \sum\limits_{n=1}^{\infty}\frac{h^{n}}{n!}$$

Thus,

$$\frac{e^{h}-1}{h} = 1 + \sum\limits_{n=2}^{\infty}\frac{h^{n}}{(n+1)!}$$

and hence tends to $1$.

EDIT: For the second part, you can express sine and cosine in terms of the exponential (Euler's formulae...)

• I don't understand why people downvoted, this answer is very clear to me. – A.E Jul 21 '13 at 13:45
• Hi, how about second part of my question(also in the last line). – HyperGroups Jul 21 '13 at 13:46
• @DonAntonio Thanks for pointing out the typo! – Vishal Gupta Jul 21 '13 at 13:57
• Fine,second part of the question maybe not a serious problem, I'll do it by definition of limit. – HyperGroups Jul 21 '13 at 13:59